Problems From JEE Adv 2026 | Paper 2 | Q9-Q14

Ajouté : 2026-05-20

[00:00:00]So, welcome back, guys. So, today we will start with problem nine.

[00:00:03]So, here we have 10 moles of an ideal monatomic gas.

[00:00:07]Uh it is initially at a state A at atmospheric pressure and temperature TA.

[00:00:11]Uh and it is enclosed in a metal cylinder of volume V0. Okay. So, we have a metal cylinder whose volume is V0.

[00:00:19]And uh initially we are at state A. And here pressure is P0, atmospheric pressure, and temperature is 300 K.

[00:00:28]And volume in this state is V0. Okay.

[00:00:31]So, it is attached to a frictionless piston. So, this piston uh is movable. It's given that the gas is suddenly compressed to state B with a volume V0 by 3. Okay. So, So, we have see a sudden compression happening and the volume of the gas is now V0 by 3. And this state is called state B. Now, uh guys, um even though the this is a metallic cylinder, which means it is like a very good conductor of heat, um and you know, as we compress the piston, the temperature of the gas, uh you know, it's going to increase. So, which means there will be heat transfer from inside to outside as the temperature of the gas rises.

[00:01:13]But, uh the thing is because this compression was done suddenly, what they are implying by saying that statement is that there wasn't enough time for any heat transfer to occur, okay, or to take place. So, which means this process going from A to B, uh we will treat it like an adiabatic process because there was not enough time for heat to escape, even though this is a metallic conductor, okay.

[00:01:39]So, this process from A to B, um is going to be an adiabatic equation. We're going to treat it like an adiabatic equation.

[00:01:46]And now we are going to keep the piston stationary. The cylinder is submerged in a water bath of temperature 11°C until the gas reaches the temperature of the water bath. Okay, so now we are going to cool it keeping the volume constant.

[00:02:02]Right? Piston is stationary. So yeah, if we if we look at A option, so A to B over here looks like an adiabatic compression process and in this process the pressure will rise. And from B to C we are keeping the volume constant and we're cooling the gas.

[00:02:17]Okay.

[00:02:18]And finally, while still in the water bath, okay, so basically in this state C, what is happening is this metallic cylinder, it was placed in a water bath and the water bath temperature was 11°C.

[00:02:33]We keep the piston in the water bath itself. Now, while the cylinder cylinder is still in the water bath, the piston is brought slowly to its initial position.

[00:02:42]Okay, which means the volume is back to V0.

[00:02:46]And this process is happening slowly.

[00:02:48]Okay.

[00:02:49]So in this last process the metallic cylinder uh it's it's slowly being expanded.

[00:02:56]Okay, this process is happening slowly.

[00:02:59]Okay, this implies that the temperature of the gas inside will be equal to the temperature of the water bath itself, which is 11°C.

[00:03:06]So this last process of this expansion happening will be an isothermal process. Okay, so this C to D will follow an isothermal process.

[00:03:16]Um and in the final state, once again, the volume is back to the original volume.

[00:03:21]Okay. So this first graph is correct and F will be below A also because in the final state the temperature of the gas is 11°C and in the initial state the temperature was 27°C, which means A will be above F in the graph, right?

[00:03:37]Okay, so this graph is actually fine.

[00:03:40]Now in B option they're saying what is the change in internal energy in going from A to B. Okay, so right to write the change in internal energy from A to B we need the temperature at A and temperature at B. So temperature at A is already known. It is 300 K and we can in order to find the temperature at B, we can use the adiabatic state equation, right? So, that will be TB B volume at B is V naught by 3. So, V naught by 3 raised to the power gamma minus 1. So, 5 by 3, this is a monatomic gas. So, 5 by 3 minus 1 will be 2 by 3 is equal to TA V naught to the power gamma minus 1.

[00:04:19]Okay, so this V naught cancels out.

[00:04:21]And the temperature at B will be temperature at A, which is 300 multiplied with 3 raised to power 2 by 3, which is 9 raised to power 1 by 3.

[00:04:31]That is given as 2.08. So, 208 times 3 So, 600 plus 24. So, 624 K is the new temperature at state B.

[00:04:42]So, the change in internal energy will be the number of moles CV into delta T, which is 324.

[00:04:51]Okay, and this after calculation will come out to be 4860R.

[00:04:56]Okay. So, option B will also be correct.

[00:04:58]Now, in C, they're saying the net change in internal energy in the whole process.

[00:05:01]So, the final state is over here, F. And the temperature at F is 11° C, which is 284 K. So, in the final So, in the the final state, the temperature is a bit less. So, delta U will be negative for the overall process. So, delta U for the whole process is uh NCV. So, the difference between 300 and 284 is negative 16, right? So, it will be minus 16. So, this will be minus 240 R. So, this is also correct.

[00:05:31]The pressure and temperature of the state B are 2.08 times the atmospheric pressure.

[00:05:36]You know, the pressure will be something else.

[00:05:39]It's the temperature that was uh 2.08 times. So, pressure will be something else, right? If you apply PV raised to power gamma.

[00:05:48]PB VB raised to power gamma is equal to P naught V naught raised to power gamma.

[00:05:57]So, here it will be something else, right? Pressure at B will be 3 to the power 5 by 3 times P naught. That is not 2.08. So, this option is wrong.

[00:06:06]Okay?

[00:06:07]So, the answer will be ABC for this.

[00:06:09]Okay, so now we have a screw gauge problem. So, we have two thin wires of diameters 0.650 and D, okay? To obtain the value of D, the diameters of the two wires are measured with a screw gauge.

[00:06:20]The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale. Okay, so if you for every 100 division rotation of the circular scale, the thimble will move forward by 0.5 mm, right? So, the smallest division on the linear scale is 0.5 mm. Okay?

[00:06:39]Uh the table shows readings of linear scale and circular scale for the measurements. Okay?

[00:06:45]So, this first measurement corresponds to the first wire and the second one corresponds to the second wire.

[00:06:50]So, So, let's calculate the first reading.

[00:06:52]So, in the first case, uh if we ignore any error and just do the calculation, it will be the linear scale reading. So, linear scale reading for wire one is 0.5 and circular scale reading is 42.

[00:07:03]So, 0.5 mm plus uh so, the number of circular scale divisions is 42.

[00:07:09]Now, per circular scale, so if we rotate the circular scale by one division, the linear scale moves by 0.5 mm divided by 100, right? So, this times 0.5 by 100 mm. So, this is 0.42.21.

[00:07:27]So, this will be 0.71 0 mm. Okay?

[00:07:32]So, this will be reading one. So, for wire one, the reading that the screw gauge is providing is 0.71, which has some error, right? So, if you want to correct it, uh we need to subtract we need to do a subtraction of So, the actual reading is 0.65. We need to subtract 0.06 mm.

[00:07:53]And then we get the true reading of 0.650.

[00:07:56]So, all we have to do is in the next reading, we just need to subtract 0.06, then we'll get the correct answer.

[00:08:04]So, for the next reading, linear scale reading is 1.5 mm >> [clears throat] >> and circular scale reading is 95 * 0.5 / 100.

[00:08:14]And in order to remove the error, we will subtract 0.06. Okay? So, this will be the answer.

[00:08:22]And this comes out to be 1.915 mm.

[00:08:27]Okay, but they asked the answer in micrometers. So, we'll just multiply this with 1,000. The answer will be 1915.

[00:08:34]Okay, so now in a single slit diffraction experiment, a slit of width So, slit width A is given with true value and error. It is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and the first minimum.

[00:08:51]Okay, so basically, if this is the central maxima, this particular angle angular width right? This is actually lambda by A.

[00:09:00]Okay, this is after using the small angle approximation, this angular width is actually lambda by A between the central maximum and the first minimum.

[00:09:08]So, [snorts] that is the angular width they're talking about and they have given it as 2° plus minus 40 arc minutes.

[00:09:17]And the value of the fractional error of the measurement of wavelength. Okay, so we'll just use the small angle approximation here.

[00:09:23]Because the angle in picture is like 2°, right? So, it is still fine.

[00:09:28]So, theta is equal to lambda by A. So, we need to find the error in fractional error in wavelength.

[00:09:35]So, delta lambda over lambda equals fractional error in theta plus fractional error in A.

[00:09:43]So, this will be equal to the error in theta. So, we need to convert this arc minute into um degrees. For that, we'll divide it with 60.

[00:09:52]>> [clears throat] >> So, this is 2/3 degrees, okay?

[00:09:55]So, the fractional error is 2/3 degrees divided by true value is 2 plus delta A is uh 0.002 divided by 0.016.

[00:10:06]So, this will be 1/3 plus this is going to be 1/8.

[00:10:13]Okay, and this after calculation turns out to be 0.4583, okay? And it repeats. So, I guess the rounded value will be 0.46.

[00:10:23]The next problem we have um a ray of ray AB of unpolarized light.

[00:10:28]Okay, and [snorts] they are showing both the planes of polarization. So, this particular line represents the oscillation of electric field in the plane of incidence, okay?

[00:10:41]And uh this particular dot represents the oscillation of the electric field into the page and out of the page, perpendicular to the plane of incidence, okay?

[00:10:50]So, okay, so clearly the incident light ray is uh unpolarized.

[00:10:56]Okay, and it undergoes some refractions like this, and it is given in the problem that at a particular incident angle I the reflected ray, which is over here, CD it is polarized. So, as you can see, here there is only one type of polarization, which is like into and out of the page. So, this is polarized light. The electric field oscillations is only along one direction rather than in multiple directions, okay? So, so this is clearly uh so clearly the Brewster's angle idea is being used here. So, as you can see over here, we have refracted light and reflected light. You know, light is incident at the Brewster's angle, then the reflected light becomes completely polarized. Now, and that's what we see over here.

[00:11:40]So, this incident angle over here is happening at the Brewster's angle.

[00:11:44]Uh and in order to find Brewster's angle, we can use the relation tan IB is refractive index of this media divided by refractive index of this media.

[00:11:54]So, 4 by root 3 divided by 4 by 3.

[00:11:58]So, this will be root 3.

[00:12:00]So, IB will turn out to be 60°.

[00:12:03]Okay, and if IB is 60°, refractive index of this media is 4 by 3.

[00:12:08]Uh and refractive index of this media is also 4 by 3, so I is also 60°.

[00:12:13]Okay.

[00:12:14]Uh for plane surface refractions, right?

[00:12:16]You can equate n sin theta at any interface. So, over here nw sin i will be equal to nw sin IB, which means IB is equal to i.

[00:12:27]So, the angle of incidence will simply be 60°. Okay, so now we have a circuit.

[00:12:33]The resistance of the galvanometer G can be found out by the half deflection method.

[00:12:39]Here, the resistance R2 is adjusted such that when this key is closed, the deflection in the galvanometer becomes half of the value as compared to when K is open.

[00:12:49]Um half deflection is obtained at R2 equal to 4 ohm. Okay. So, at the start, uh of course, the current that was flowing in the circuit was IG um when the key was open. So, we can write IG as EMF divided by total resistance, which is R1 plus galvanometer resistance is 6 ohm.

[00:13:11]Okay. Now, when the key is closed, uh and they have given R2 value as 4 ohm.

[00:13:17]And the galvanometer resistance is 6 ohm.

[00:13:20]They mentioned that this is the half deflection case. So, So, what that means is when IG was flowing the deflection was full.

[00:13:29]Basically the max current reading was being shown, right? And when we closed K, the deflection became half. So now it is max by two.

[00:13:40]So [snorts] this deflection is directly proportional to the current flowing. So the new current flowing through the galvanometer branch is IG by two.

[00:13:48]Now because the resistances are different, the current here will be something else.

[00:13:55]So that we can find it out. So IG by two into six is the voltage across the galvanometer. This will be equal to the current in the four ohm resistor times four. So current in the four ohm resistor will be IG into three by four. Three by four IG, right?

[00:14:14]So this current is three by four IG.

[00:14:18]So the total current in the circuit will be three by four plus half which is five by four IG.

[00:14:27]So in the second case the same equation as this will be five by four IG equals the EMF divided by equivalent resistance. So equivalent resistance will be R1 plus six and four the equivalent will be six into four by 10.

[00:14:43]So 2.4.

[00:14:44]So R1 plus 2.4 ohm.

[00:14:48]So now we can just use these two equations. So five by four into IG which is 10 by R1 plus six equals 10 divided by R1 plus 2.4. So 10 cancels out.

[00:15:02]So five R1 2.4 times five is 12 equals four R1 plus 24.

[00:15:12]So R1 turns out to be 12 ohms.

[00:15:15]Then IG will be 10 divided by 18.

[00:15:19]So IG is going to be 10 by 18 amp amperes.

[00:15:25]Uh the question is what is the current in the half deflection case through R1?

[00:15:29]So, that will be 5 by 4 IG.

[00:15:32]So, the answer is 5 by 4 into 10 by 18 amperes. So, 25 by 36, but I guess the answer they wanted in milliampere. So, we'll multiply with 1,000.

[00:15:46]Okay, so this will come out to be 694.44 milliamperes. Okay.

[00:15:53]Okay, so the next problem is a very simple units and dimension question.

[00:15:57]So, we have a system of units.

[00:15:59]A new system of units in which mass, length, time, and current are made five x basically. So, then the question is what happens to the magnitude of one SI unit of this particular quantity root mu naught by epsilon naught. Okay, so here we'll first find what is the units of root mu naught by epsilon naught. Okay, so for finding out the SI unit of this, the dimension of this quantity or the unit of this quantity, uh what we can do is we can first write this as root mu naught epsilon naught divided by epsilon naught. Okay. So, what this will do is this root mu naught epsilon naught we can write it in terms of speed. Okay. One by speed of light can be expressed as one by root mu naught epsilon naught. So, this was actually the formula for the speed of light, but it's but the reciprocal. Okay.

[00:16:49]So, the unit of one by root mu naught epsilon naught is uh one by the unit of speed, which is meters per second. Okay.

[00:16:57]Now, we just need the unit of epsilon naught, which we can use uh Coulomb's law to see. So, one by epsilon naught charge uh Q squared. So, Q squared we can write it as So, Q squared we can write it as ampere squared into second squared.

[00:17:12]And uh divided by distance squared is meter squared.

[00:17:16]This was force. So, force was uh kg m per second square, right? So, epsilon not's unit is A squared s to the power 4 divided by kg m cubed, okay? So, now we can just substitute that over here.

[00:17:38]So, root mu naught by epsilon naught, its unit is s by m and over here we have A squared s to the power 4 divided by kg m cubed. Okay.

[00:17:51]So, this will become kg. So, this is 1 kg m squared divided by uh A squared * s cubed, okay? So, this is the unit of root of mu naught by epsilon naught.

[00:18:08]Uh now we are scaling this, you know?

[00:18:10]In the original SI system, the SI unit was 1 kg, 1 m the whole square, 1 A the whole square, and 1 s the whole cube, right? So, now we are scaling that to 5 kg, 5 m, 5 s, and 5 A. Okay.

[00:18:23]So, this is also equal to So, what we can do is instead of the original 1 kg unit, now it will be 5 kg.

[00:18:31]Uh so, if you're multiplying with 5, we have to divide with 5 as well. Um so that these two quantities are equal, right?

[00:18:39]So, now the unit of meter is the new unit of meter is 5 m.

[00:18:45]5 m squared, so we are multiplying it with 5 squared, so divide it with 5 squared as well.

[00:18:50]Now, the new unit of A is 5 A.

[00:18:53]So, we are dividing with 5 squared, so we'll multiply with 5 squared as well.

[00:18:57]And uh similarly, we are making the new unit of second is 5 s. There's a cube here.

[00:19:02]So, we'll multiply with 5 cubed as well, okay? So, this over here is the new new unit. And this over here is the magnitude in this system of units, so that will be 25.

[00:19:13]25 * this new unit, okay?

[00:19:17]25 * this new unit is equal to 1 * the original SI unit. So, the question was what is this new magnitude? So, the answer will be 25.

[00:19:27]All right?

[00:19:28]So, that is it, guys. That is it for this video. We'll meet in the next video.